Modern Engineering Thermodynamics

224 CHAPTER 7: Second Law of Thermodynamics and Entropy Transport and Production Mechanisms
EXAMPLE 7.4 (Continued )
Insulation
FIGURE 7.12
Example 7.4.
m = 1.5 kg
x 1 = 0
T 1 = 20.°C
p 1 = 0.10 MPa
s 1 = ?
State 1
Constant volume process
State 2
s 2 − s 1 = ?
m = 1.5 kg
p 2 = 0.10 MPa
∀ 2 =∀ 1
s 2 = ?
or, T 2 = T 1 . Therefore, the process must also be isothermal, and Eq. (7.33) gives the specific entropy change as
s 2 − s 1 = c lnðT 2 =T 1 Þ = c lnð1Þ = 0
Consequently, the entropy of an incompressible material is not altered by changing its pressure.
EXAMPLE 7.5
An apparatus contains 0.035 kg of air (an ideal gas). The apparatus is used to compress the air isentropically (i.e., at
constant entropy) from a pressure of 0.100 MPa to a pressure of 5.00 MPa. If the initial temperature of the air is 20.0°C,
determine the final temperature and specific volume of the air.
Solution
First, draw a sketch of the system (Figure 7.13).
Air
State 1
m = 0.035 kg
p 1 = 0.100 MPa
T 1 = 20.0°C
Isentropic process
State 2
Air
m = 0.035 kg
p 2 = 5.00 MPa
T 2 = ?
v 2 = ?
FIGURE 7.13
Example 7.5.
The unknowns are the final temperature, T 2 , and specific volume, v 2 , of the air in the system. Since p 1 = 0.100 MPa,
T 1 = 20.0°C, and p 2 = 5.00 MPa, then Eq. (7.38) can be used to find T 2 and v 2 as follows. From Thermodynamic Tables to
accompany Modern Engineering Thermodynamics, Table C.13b, we find for air that k = 1.4; and solving Eq. (7.38) for T 2 gives
k−1
p
T 2
2 = T
k
5:00 MPa
0:4
1 = ð20:0 + 273:15 KÞ
1:4 = 896 K = 623°C
p 1
0:100 MPa
Using the ideal gas equation of state and Table C.13b for the gas constant of air (R air = 0.286 kJ/kg · K), we find the initial
specific volume of the air to be
v 1 = mRT 1 /p 1
= ð0:035 kgÞð0:286 kJ=kg.KÞð20:0 + 273:15 KÞ/ð0:100 × 10 3 kN=m 2 Þ
= 0:02934 m 3 =kg
Then, solving Eq. (7.38) for v 2 gives
1
1 −
T
v 2 = v 2 1−k
1 = ð0:02934 m 3 623 + 273:15
=kgÞ
0:4
= 0:00180 m 3 =kg
T 1
20:0 + 273:15
Exercises
7. Determine the change in specific entropy as a 1.00 kg block of solid incompressible iron is heated from 20.0°C to 100.°C.
(See Table 3.6 for specific heat values.) Answer: (s 2 − s 1 ) iron = 0.108 kJ/kg·K.
8. Determine the change in specific entropy of air as it is heated from 20.0°C to 100.°C in a constant pressure (isobaric)
process. Assume air behaves as an ideal gas. Answer: (s 2 − s 1 ) isobaric air = 0.242 kJ/kg·K.
9. Determine the change in specific entropy of air as it is heated from 20.0°C to 100.°C in a constant volume (isochoric)
process. Assume air behaves as an ideal gas. Answer: (s 2 − s 1 ) isochoric air = 0.173 kJ/kg·K.